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参数疏导：预计 MSE
统计学问题：级别(k\)修剪后的平均值的MSE是多少？
咱们如何答复它：预计从规范柯西散布（t 散布 w/df = 1）生成的大小为 20 的随机样本的程度 \(k\) 修剪均值的 MSE。指标参数 \(\theta\) 是核心或中位数。柯西散布不存在均值。在表中总结 MSE 的估计值 \(k = 1, 2, ... 9\)。
result=rep(0,9)for(j in 1:9){ n<-20 for(i in 1:m){ x<-sort(rcauchy(n))参数自抽样法：教训效用计算
统计问题：随着零假如与事实之间的差别发生变化，效用如何变动？
咱们如何答复：绘制 t 测验的教训效用曲线。
t 测验的原假如是 。另一种抉择是。
您将从具备 的正态分布总体中抽取大小为 20 的样本。您将应用 0.05 的显着性程度。
显示当总体的理论平均值从 350 变为 650（增量为 10）时，效用如何变动。
y 轴是教训效用（通过 bootstrap 预计），x 轴是 \(\mu\) 的不同值（350、360、370 … 650）。
x <- rnorm(n, mean = muA, sd = sigma) #抽取平均值=450的样本 ts <- t.test(x, mu = mu0) #对有效的mu=500进行t测验 ts$p.value
参数自抽样法：教训效用计算
统计问题：样本量如何影响效用？
咱们如何答复：创立更多的效用曲线，因为理论均值在 350 到 650 之间变动，但应用大小为 n = 10、n = 20、n = 30、n = 40 和 n = 50 的样本生成它们。同一图上的所有 5 条效用曲线。
pvals <- replicate(m, pvalue())power <- mean(pvals <= 0.05)points(sequence,final2\[2,\],col="red",pch=1)points(sequence,final2\[3,\],col="blue",pch=2)
参数自抽样法：教训置信水平
统计问题：在制作 95% CI 时，如果咱们的样本很小并且不是来自正态分布，咱们是否仍有 95% 的置信度？
咱们如何答复它：依据样本为总体的平均值创立一堆置信区间 (95%)。
您的样本大小应为 16，取自具备 2 个自由度的卡方散布。
找出未能捕获总体实在均值的置信区间的比例。（揭示：自由度为 \(k\) 的卡方散布的平均值为 \(k\)。）
for(i in 1:m){ samp=rchisq(n,df=2) mean=mean(samp) sd=sd(samp) upper=mean+qt(0.975,df=15)*sd/4
非参数自抽样法置信区间
统计问题：基于一个样本，咱们能够为总体相关性创立一个置信区间吗？
咱们如何答复：为相干统计量创立一个 bootstrap t 置信区间预计。
boot.ti <- function(x, B = 500, R = 100, level = .95, stattic){ x <- as.matrix(x) library(boot) #for boot and boot.cidata(law, package = "bootstrap")dat <- lawci <- boot.t.ci(dat, statistic = stat, B=2000, R=200)ci
自抽样法后的Jackknife
统计问题：R 的标准误差的 bootstrap 预计的标准误差是多少？
咱们如何答复它： data(law) 像上一个问题一样应用。在 bootstrap 后执行 Jackknife 以取得标准误差预计的标准误差预计。（bootstrap 用于取得总体中 R 的 SE 的估计值。而后应用折刀法取得该 SE 估计值的 SE。）
indices <- matrix(0, nrow = B, ncol = n)# 进行自举for(b in 1:B){ i <- sample(1:n, size = n, replace = TRUE) LSAT <- law$LSAT\[i\] # jackknifefor(i in 1:n){ keepers <- function(k){ !any(k == i) }
自测题
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Parametric Bootstrap: Estimate MSE
Statistical question: What is the MSE of a level \(k\) trimmed mean?
How we can answer it: Estimate the MSE of the level \(k\) trimmed mean for random samples of size 20 generated from a standard Cauchy distribution (t-distribution w/df = 1). The target parameter \(\theta\) is the center or median. The mean does not exist for a Cauchy distribution. Summarize the estimates of MSE in a table for \(k = 1, 2, ... 9\).
Parametric Bootstrap: Empirical Power Calculations
Statistical question: How does power change as the difference between the null hypothes and the reality changes?
How we can answer it: Plot an empirical power curve for a t-test.
The null hypothesis of the t-test is \(\mu = 500\). The alternative is \(\mu \ne 500\).
You will draw samples of size 20, from a normally distributed population with \(\sigma = 100\). You will use a significance level of 0.05.
Show how the power changes as the actual mean of the population changes from 350 to 650 (increments of 10).
On the y-axis will be the empirical power (estimated via bootstrap) and the x-axis will be the different values of \(\mu\) (350, 360, 370 … 650).
Parametric Bootstrap: Empirical Power Calculations
Statistical question: How does sample size affect power?
How we can answer it: Create more power curves as the actual mean varies from 350 to 650, but produce them for using samples of size n = 10, n = 20, n = 30, n = 40, and n = 50. Put all 5 power curves on the same plot.
Parametric Bootstrap: Empirical Confidence Level
Statistical question: When making a 95% CI, are we still 95% confident if our samples are small and do not come from a normal distribution?
How we can answer it: Create a bunch of Confidence Intervals (95%) for the mean of a population based on a sample.
\[\bar{x} \pm t^{*} \times \frac{s}{\sqrt{n}}\]
Your samples should be of size 16, drawn from a chi-squared distribution with 2 degrees of freedom.
Find the proportion of Confidence Intervals that fail to capture the true mean of the population. (Reminder: a chi-squared distribution with \(k\) degrees of freedom has a mean of \(k\).)
Non Parametric Bootstrap Confidence Interval
Statistical question: Based on one sample, can we create a confidence interval for the correlation of the population?
How we can answer it: Create a bootstrap t confidence interval estimate for the correlation statistic.
Jackknife after bootstrap
Statistical question: What is the standard error of the bootstrap estimate of the standard error of R?
How we can answer it: Use data(law) like the previous problem. Perform Jackknife after bootstrap to get a standard error estimate of the standard error estimate. (The bootstrap is used to get an estimate of the SE of R in the population. The jackknife is then used to get an SE of that SE estimate.)
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